Special case of this general integral












2












$begingroup$


We know that the integral of,



$$ int frac{1}{x} dx = ln x$$



ignoring any integration constants.



Consider the integral of some more `general' function,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$



i.e. in the $y=z=0$ case we recover the original integral ($1/x$)



Now, if I put the second integral into WolframAlpha, it spits out,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.



I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$



What gives?






integration







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 18:30
















2












$begingroup$


We know that the integral of,



$$ int frac{1}{x} dx = ln x$$



ignoring any integration constants.



Consider the integral of some more `general' function,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$



i.e. in the $y=z=0$ case we recover the original integral ($1/x$)



Now, if I put the second integral into WolframAlpha, it spits out,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.



I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$



What gives?






integration







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 18:30














2












2








2





$begingroup$


We know that the integral of,



$$ int frac{1}{x} dx = ln x$$



ignoring any integration constants.



Consider the integral of some more `general' function,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$



i.e. in the $y=z=0$ case we recover the original integral ($1/x$)



Now, if I put the second integral into WolframAlpha, it spits out,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.



I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$



What gives?






integration







share|cite|improve this question









$endgroup$




We know that the integral of,



$$ int frac{1}{x} dx = ln x$$



ignoring any integration constants.



Consider the integral of some more `general' function,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx $$



i.e. in the $y=z=0$ case we recover the original integral ($1/x$)



Now, if I put the second integral into WolframAlpha, it spits out,



$$ int frac{1}{sqrt{x^2 + y^z +z^2}} dx = ln left(sqrt{x^2+y^2+z^2} + xright)$$.



I naively expected that in the $y=z=0$ case, this answer would revert to $ln x$, but instead we get $ln 2x$



What gives?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 18:25









user1887919user1887919

225112




225112








  • 4




    $begingroup$
    Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 18:30














  • 4




    $begingroup$
    Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 18:30








4




4




$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30




$begingroup$
Technically it's $intfrac{1}{x}dx=ln|x|+C$, where the locally constant function $C$ can have different values for the cases $x<0,,x>0$.
$endgroup$
– J.G.
Dec 4 '18 at 18:30










1 Answer
1






active

oldest

votes


















6












$begingroup$

Note that



$$ln 2x=ln x +ln 2=ln x +C$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
    $endgroup$
    – user1887919
    Dec 4 '18 at 18:30










  • $begingroup$
    @user1887919 No since the constant cancels out, let try also with some numerical example.
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:31












  • $begingroup$
    @user1887919 You are welcome! Refer also to that related OP, Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Note that



$$ln 2x=ln x +ln 2=ln x +C$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
    $endgroup$
    – user1887919
    Dec 4 '18 at 18:30










  • $begingroup$
    @user1887919 No since the constant cancels out, let try also with some numerical example.
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:31












  • $begingroup$
    @user1887919 You are welcome! Refer also to that related OP, Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:36
















6












$begingroup$

Note that



$$ln 2x=ln x +ln 2=ln x +C$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
    $endgroup$
    – user1887919
    Dec 4 '18 at 18:30










  • $begingroup$
    @user1887919 No since the constant cancels out, let try also with some numerical example.
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:31












  • $begingroup$
    @user1887919 You are welcome! Refer also to that related OP, Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:36














6












6








6





$begingroup$

Note that



$$ln 2x=ln x +ln 2=ln x +C$$






share|cite|improve this answer









$endgroup$



Note that



$$ln 2x=ln x +ln 2=ln x +C$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 18:27









gimusigimusi

92.9k94494




92.9k94494












  • $begingroup$
    Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
    $endgroup$
    – user1887919
    Dec 4 '18 at 18:30










  • $begingroup$
    @user1887919 No since the constant cancels out, let try also with some numerical example.
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:31












  • $begingroup$
    @user1887919 You are welcome! Refer also to that related OP, Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:36


















  • $begingroup$
    Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
    $endgroup$
    – user1887919
    Dec 4 '18 at 18:30










  • $begingroup$
    @user1887919 No since the constant cancels out, let try also with some numerical example.
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:31












  • $begingroup$
    @user1887919 You are welcome! Refer also to that related OP, Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 18:36
















$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30




$begingroup$
Hmmm ok, but if I wanted to evaluate both over some limits $x=x_1 rightarrow x_2$, it seems I would still get different answers?
$endgroup$
– user1887919
Dec 4 '18 at 18:30












$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31






$begingroup$
@user1887919 No since the constant cancels out, let try also with some numerical example.
$endgroup$
– gimusi
Dec 4 '18 at 18:31














$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36




$begingroup$
@user1887919 You are welcome! Refer also to that related OP, Bye
$endgroup$
– gimusi
Dec 4 '18 at 18:36


















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